I'm just trying to understand the concept of the adjoint of unbounded operators.
Let's look at the operator $Au:=-\epsilon\Delta u +(b\cdot\nabla)u+cu$
as an unbounded Operator on $L^{2}(\Omega)$ with $\epsilon>0$ and $b,c\in L^{\infty}(\Omega)$.
If we set $D(A):=H^{1}_{0}(\Omega)\cap H^{2}(\Omega)$ and the boundary of $\Omega$ is sufficently smooth, $A$ is a densely defined, closed operator.
Now, obviously the formal adjoint $F(A)$ of $A$ is $$F(A)u:=-\epsilon\Delta u -(b\cdot\nabla)u+cu$$ and we can give it the same Domain as $A$, i.e. $H^{1}_{0}(\Omega)\cap H^{2}(\Omega)$ to make it a densely defined closed Operator on $L^{2}(\Omega)$ with $$(Ax,y)_{L^{2}(\Omega)}=(x,F(A)y)_{L^{2}(\Omega)}$$ for all $x,y\in D(A)=D(F(A))$.
Now my question is: Is $A^{*}=F(A)$ with these definitions ? If so, why ? Why arent there any $y\in L^{2}(\Omega)\setminus D(A)$ so that $$(Ax,y)_{L^{2}(\Omega)}\leq K_{y} \|x\|_{L^{2}(\Omega)}$$ for all $x\in D(A)$ ?
Edit: I think I found an answer, though im not 100% convinced by it, because it seems to circumvent the condition $$y\in D(A^{*}) \Leftrightarrow (Ax,y)_{L^{2}(\Omega)}\leq K_{y} \|x\|_{L^{2}(\Omega)}\;\;\text{for all}\;\;x\in D(A)$$
Assertion: $D(A^{*})=D(A)$.
Proof: With the above mentioned condition and some integration by parts, it is easy to see, that $D(A)\subset D(A^{*})$. Let's show the other inclusion. For this we recall from standard theory of elliptic equations, that there exists a $\mu\in\mathbb{R}$, so that for all $\lambda\geq\mu$ the map $$A+\lambda I:D(A)\rightarrow L^{2}(\Omega)$$ is an isomorphism. Since $D(A+\lambda I)=D(A)$, we can assume from now on, that $A$ is a bijection. This is really the whole idea. On $D(A)$ we have $F(A)=A^{*}$ so $A^{*}$ is surjective. It is easy to see that $$\text{ker}(A^{*})=\text{im}(A)^{\bot}=L^{2}(\Omega)^{\bot}=\{0\}$$ so $A^{*}$ is also injective. If there was a $y\in L^{2}(\Omega)\setminus D(A)$ with $y\in D(A^{*})$, we would have an $x\in L^{2}(\Omega)$ and a $\hat{y}\in D(A)$ with
$$A^{*}y=x=A^{*}\hat{y}$$ rendering $A^{*}$ not injective and proving the assertion by contradiction. $\square$
I am sorry about having posed this question and then shortly after coming up with an answer myself. I really did think about the problem for quite a few hours. I guess it helped me to formally explain my problem to others. If anyone finds a problem with my proof, please tell me ! I guess the updated question is: Is this proof correct ?